ko có GT thỏa mãn

a=0

Do: A=1^4+2.1^2-1^3+1-0=0

B=1-1=0. mà 0 thì chia hết cho 0.

Đặt \(P=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)
\(P=x^4\left(y-z\right)+y^4\left(z-x\right)+z^4\left(x-y\right)\)

\(=x^4\left(y-z\right)+y^4z-y^4x+z^4x-z^4y\)

\(=x^4\left(y-z\right)+y^4z-z^4y-y^4x+z^4x\)

\(=x^4\left(y-z\right)+yz\left(y^3-z^3\right)-x\left(y^4-z^4\right)\)

\(=x^4\left(y-z\right)+yz\left(y-z\right)\left(y^2+yz+z^2\right)-x\left(y-z\right)\left(y^3+y^2z+yz^2+z^3\right)\)

\(=\left(y-z\right)\left[x^4+yz\left(y^2+yz+z^2\right)-x\left(y^3+y^2z+yz^2+z^3\right)\right]\)

\(=\left(y-z\right)\left(x^4+y^3z+y^2z^2+yz^3-xy^3-xy^2z-xyz^2-xz^3\right)\)

\(=\left(y-z\right)\left(x^4-xz^3-xy^3+y^3z-xy^2z+y^2z^2-xyz^2+yz^3\right)\)

\(=\left(y-z\right)\left[x\left(x^3-z^3\right)-y^3\left(x-z\right)-y^2z\left(x-z\right)-yz^2\left(x-z\right)\right]\)

\(=\left(y-z\right)\left[x\left(x-z\right)\left(x^2+xz+z^2\right)-y^3\left(x-z\right)-y^2z\left(x-z\right)-yz^2\left(x-z\right)\right]\)

\(=\left(y-z\right)\left(x-z\right)\left[x\left(x^2+xz+z^2\right)-y^3-y^2z-yz^2\right]\)

\(=\left(y-z\right)\left(x-z\right)\left(x^3+x^2z+xz^2-y^3-y^2z-yz^2\right)\)

\(=\left(y-z\right)\left(x-z\right)\left(x^3-y^3+x^2z-y^2z+xz^2-yz^2\right)\)

\(=\left(y-z\right)\left(x-z\right)\left[\left(x-y\right)\left(x^2+xy+y^2\right)+z\left(x^2-y^2\right)+z^2\left(x-y\right)\right]\)

\(=\left(y-z\right)\left(x-z\right)\left[\left(x-y\right)\left(x^2+xy+y^2\right)+z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\right]\)

\(=\left(y-z\right)\left(x-z\right)\left(x-y\right)\left[x^2+xy+y^2+z\left(x+y\right)+z^2\right]\)

\(=\left(y-z\right)\left(x-z\right)\left(x-y\right)\left(x^2+xy+y^2+xz+yz+z^2\right)\)

Đặt \(A=x^2+xy+y^2+xz+yz+z^2\)

\(A=\frac{2\left(x^2+xy+y^2+xz+yz+z^2\right)}{2}=\frac{2x^2+2xy+2y^2+2xz+2yz+2z^2}{2}\)

\(=\frac{\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(x^2+2xz+z^2\right)}{2}\)

\(=\frac{\left(x+y\right)^2+\left(y+z\right)^2+\left(x+z\right)^2}{2}\)

=>\(P=\left(y-z\right)\left(x-z\right)\left(x-y\right).\frac{\left(x+y\right)^2+\left(y+z\right)^2+\left(x+z\right)^2}{2}\)

Ta có: \(x>y>z< =>\hept{\begin{cases}x>y\\y>z\\x>z\end{cases}}< =>\hept{\begin{cases}x-y>0\\y-z>0\\x-z>0\end{cases}}\)

Dễ thấy \(\left(x+y\right)^2\ge0;\left(y+z\right)^2\ge0;\left(x+z\right)^2\ge0\) với mọi x;y;z

\(=>P>0\) (đpcm)

vì x^200 chia hết cho 4 , x^100 chia hết cho x^2 và 1 chia hết cho 1 nên x^200+x^100+1 chia hếtcho x^4+x^2+1

**** bn nhe  

Đặt x2=ax2=a. Cần chứng minh: a^100+a^50⋮a2+a+1a100+a50⋮a2+a+1

Sử dụng tính chất quen thuộc: a3m+1+a3n+2=a(a3m−1)+a2(a3n−1)−(a2+a+1)⋮a2+a+1

ta có: đặt A là f(x) và B là g(x)

f(x) chia hết cho g(x) khi f(x) = g(x).q(x)

<=> x⁴ + 2x² - x³ + x - a = (x - 1) . q(x)

<=> (1)⁴ + 2(1)² - (1)³ + 1 - a = (1  - 1) . q(x)

<=> 4 + 2 - 1 + 1 - a = 0

<=> 6 - a = 0

<=> a = 6

ok mk nhá!! 4353453463645745756856856345345634645764767657878732545345245245665465476

bài này nếu mk ko nhầm thì là bài tự luận chương tứ giác của bài giảng lp 8,có lẽ bn nên tự làm hay hơn

Gợi ý:Áp dụng đ/lý tổng các góc trong 1 tứ giác và tận dụng triệt để các gt về tia p/giác

c) xét giá trị riêng

\(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\)

\(=xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+xyz+xyz\)

\(=xy\left(x+y\right)+y^2z+yz^2+x^2z+xz^2+xyz+xyz\)

\(=xy\left(x+y\right)+y^2z+xyz+yz^2+xz^2+x^2z+xyz\)

\(=xy\left(x+y\right)+yz\left(x+y\right)+z^2\left(x+y\right)+xz\left(x+y\right)\)

\(=\left(x+y\right)\left(xy+yz+z^2+xz\right)\)

\(=\left(x+y\right)\left[y\left(x+z\right)+z\left(x+z\right)\right]=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

a) \(x^2-y^2-x-y\)
\(=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)

Câu B tương tự nha :

 \(\left(x-y+z\right)^2+\left(z-y\right)^2+\left(x-y+z\right)\left(2z-2y\right)\)

\(=\left(x-y+z\right)^2-2\left(z-y\right)\left(x-y+z\right)+\left(z-y\right)^2\)

\(=\left(x-y+z-z+y\right)^2\)

\(=x^2\)

câu b nha  ( a + b )( a ^ 2 - ab + b ^ 2 ) -( a - b )( a ^ 2 + ab + b ^ 2 ) = (a^3 - a^2 * b + ab^2 + ba^2 - ab^2 + b^3)  - (a^3 + a^2 * b + ab^2  - a^2 * b - ab^2 - b^3) = (a^3 + b^3 ) - (a^3 - b^3) = 2b^3 

Ta có:  (a³+a²b+ab²+b³)(a-b)=a⁴-b⁴

=> a⁴+a³b+a²b²+ab³-a³b-a²b²-ab³-b⁴=a⁴-b⁴

=> (a³b-a³b)+(a²b²-a²b²)+(ab³-ab³)+(a⁴-b⁴)= a⁴-b⁴

=> a⁴-b⁴=a⁴-b⁴

=>  ĐPCM
 

Xét vế trái

\(\left(a^3+a^2b+ab^2+b^3\right)\left(a-b\right)\)

\(=a^4+a^3b+a^2b^2+ab^3-a^3b-a^2b^2-ab^3-b^4\)

\(=a^4-b^4\)

= vế phải 

=> Đpcm